Question

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10¯¹ N m¯¹. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.

Answer 1

Given, the radius of the mercury drop is 3.0 mm, the surface tension of mercury at 20 °C is 4.65× 10 −1  N/m , and the atmospheric pressure is 1.01× 10 5  Pa.

The excess pressure inside the drop of mercury is given by the equation,

P 1 = 2S r

Substituting the values in the above equation, we get:

P 1 = 2×4.65× 10 −1  N/m 3.0× 10 −3  m =310 N/ m 2

Let P 0 be the atmospheric pressure.

The total pressure inside the drop of mercury is given by the equation,

P= P 0 + P 1 =1.01× 10 5  Pa+310 N/ m 2 =1.0131× 10 5  Pa

Hence, the total pressure inside the drop of mercury is 1.0131× 10 5  Pa.